Properties of Indices (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Laws of Indices

What laws of indices do I need to know?

In an expression like $a^{n}$
- $a$ is known as the base
- $n$ is known as the index (also called the power or exponent)
The index laws you need to know and be able to use are summarised here:
- $a^{m} \times a^{n} = a^{m + n}$
- $a^{m} \div a^{n} = a^{m - n}$
  - Be careful! The two laws above can only be used if the two terms on the left-hand side of the equation have the same base.
- ${(a^{m})}^{n} = a^{m n}$
- ${(a b)}^{n} = a^{n} b^{n}$
- $a^{1} = a$
- $a^{0} = 1$
- $a^{\frac{1}{m}} = \sqrt[m]{a}$
- $a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = {(\sqrt[n]{a})}^{m}$
- $a^{- m} = \frac{1}{a^{m}}$

How do I work with laws of indices?

Laws of indices work with numerical and algebraic terms
These can be used to simplify expressions where terms are multiplied or divided
- Deal with the number and algebraic parts separately
  - $(3 x^{7}) \times (6 x^{4}) = (3 \times 6) \times (x^{7} \times x^{4}) = 18 x^{11}$
  - $\frac{3 x^{7}}{6 x^{4}} = \frac{3}{6} \times \frac{x^{7}}{x^{4}} = \frac{1}{2} x^{3}$
  - ${(3 x^{7})}^{2} = {(3)}^{2} \times {(x^{7})}^{2} = 9 x^{14}$

How can I solve equations when the unknown is in the index?

If two terms with indices are equal and the terms have the same positive base (other than 1) then the indices must be equal
- If $a^{x} = a^{y}$ then $x = y$
  - Not valid if $a \leq 0$ or $a = 1$
If the unknown is part of the index then write both sides with the same base number
- Then ignore the base number, make the indices equal and solve that equation

$\begin{array}{rcl} 5^{2 x} & = & 125 \\ 5^{2 x} & = & 5^{3} \\ 2 x & = & 3 \\ x & = & \frac{3}{2} \end{array}$

In more complicated questions you might have to use negative and fractional indices
- You may also have to rewrite both sides with the same base number

$\begin{array}{rcl} 8^{x} & = & \frac{1}{4} \\ {(2^{3})}^{x} & = & \frac{1}{2^{2}} \\ 2^{3 x} & = & 2^{- 2} \\ 3 x & = & - 2 \\ x & = & - \frac{2}{3} \end{array}$

Worked Example

(a) $\begin{matrix} ^{} \end{matrix}$ Simplify $\frac{(3 x^{2}) (2 x^{3} y^{2})}{(6 x^{2} y)}$ .

Multiply out the brackets in the numerator.
Rearrange the numerator so that you are multiplying the numbers together, the $x$ terms together and the $y$ terms together.

$\frac{3 \times 2 \times x^{2} \times x^{3} \times y^{2}}{6 x^{2} y}$

Simplify the numerator.
Multiply the constants together and add the powers of the $x$ terms together.

$\frac{6 x^{5} y^{2}}{6 x^{2} y}$

Divide the constants.
Subtract the power of the $x$ term in the denominator from the $x$ term in the numerator: $x^{5 - 2} = x^{3}$ .
Subtract the power of the $y$ term in the denominator from the $y$ term in the numerator: $y^{2 - 1} = y^{1}$ .

$x^{3} y$

(b) Simplify ${(\frac{54 x^{7}}{2 x^{4}})}^{- \frac{1}{3}}$ .

Simplify the expression inside the brackets.
Cancel down the constants.
Subtract the power of the $x$ term in the denominator from the $x$ term in the numerator: $x^{7 - 4} = x^{3}$ .

${(27 x^{3})}^{- \frac{1}{3}}$

Apply the negative index outside the brackets by 'flipping' the fraction inside the brackets.

${(\frac{1}{27 x^{3}})}^{\frac{1}{3}}$

Apply the fractional index outside the brackets to everything inside the brackets.

$\frac{1^{\frac{1}{3}}}{27^{\frac{1}{3}} x^{3 \times \frac{1}{3}}}$

Simplify.

$\frac{1}{3 x}$

Last updated: 16 October 2024

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